I calculate both u'(x) and u"(x) to be equal to Ae^x. I dont know how the A^2 appears in front of u(x)!!!.....

Thanks

Dec 3rd 2010, 04:30 AM

HallsofIvy

Why do you say that $\displaystyle u''(x)$ "is calculated as $\displaystyle u''(x)= A^2u(x)$? If $\displaystyle u(x)= Ae^x$ then $\displaystyle u'= Ae^x$ and $\displaystyle u''= Ae^x$. That can be written as $\displaystyle u''= k^2u(x)$ if and only if $\displaystyle k^2= A$ so that $\displaystyle k= \sqrt{A}$.

Dec 3rd 2010, 06:52 AM

bugatti79

My lecture notes reads
$\displaystyle u"(x)=k^2(x)$

$\displaystyle u(x)=Ae^x$ therefore

$\displaystyle u"(x)=(u'(x))'=(Au(x))'=Au'(x)=A^2u(x)$ ????

Therefore $\displaystyle A^2u(x)=k^2u(x)$ implies

$\displaystyle A_1=k and A_2=-k$

I dont understand the third line!!...I dont think its the standard way of getting solution for a nd order equation?
Thanks

Dec 3rd 2010, 07:05 AM

hmmmm

so you have $\displaystyle u'(x)=Au(x)$ and $\displaystyle u(x)=Ae^x$ ?? In the third line, that seems strange?

Dec 3rd 2010, 09:16 AM

bugatti79

I dont know what his method is. I think I will just stick with the standard way of getting the solution ie,

Why did you change from k to n? If the differential equation is $\displaystyle u''= k^2u$, then the characteristic equation is $\displaystyle m^2= k^2$ so that $\displaystyle m= \pm k$ and the general solution to the differential equation is $\displaystyle u(x)= Ae^{kx}+ Be^{-kx}$. Note that this could also be written as $\displaystyle u(x)= Ccosh(kx)+ Dsinh(kx)$.

By the way, using two consecutive single quotes in LaTex rather than a double quote will give $\displaystyle u''$ rather than $\displaystyle u"$.

Dec 5th 2010, 10:46 AM

bugatti79

Quote:

Originally Posted by HallsofIvy

Why did you change from k to n? If the differential equation is $\displaystyle u''= k^2u$, then the characteristic equation is $\displaystyle m^2= k^2$ so that $\displaystyle m= \pm k$ and the general solution to the differential equation is $\displaystyle u(x)= Ae^{kx}+ Be^{-kx}$. Note that this could also be written as $\displaystyle u(x)= Ccosh(kx)+ Dsinh(kx)$.

By the way, using two consecutive single quotes in LaTex rather than a double quote will give $\displaystyle u''$ rather than $\displaystyle u"$.